A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1}$ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?
0·3 m s$^{-1}$
3 m s$^{-1}$
−3 m s$^{-1}$
−0·3 m s$^{-1}$
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC NDA-2 – 2022
– The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest.
– According to the conservation of momentum, the total momentum of the system after firing must also be zero.
– Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol.
– Momentum before firing = 0
– Momentum after firing = $m_b v_b + m_p v_p$
– By conservation of momentum: $m_b v_b + m_p v_p = 0$
– Given: $m_b = 10 \text{ g} = 0.01 \text{ kg}$, $v_b = 300 \text{ m/s}$, $m_p = 1 \text{ kg}$.
– Substituting the values: $(0.01 \text{ kg})(300 \text{ m/s}) + (1 \text{ kg}) v_p = 0$
– $3 \text{ kg} \cdot \text{m/s} + v_p \text{ kg} = 0$
– $v_p = -3 \text{ m/s}$